Work done by a constant force

When a force acts on an object through a displacement, it transfers energy. The energy transferred by the force is called work done. For a constant force, the work done depends on the size and direction of the force and the object’s displacement:

Here, $$W$$ is work done, $$F$$ is the force, $$s$$ is the displacement, and $$\theta$$ is the angle between the force and the displacement. Work is measured in joules: $$1,\mathrm{J} = 1,\mathrm{N,m}$$.

The term $$F\cos\theta$$ is the component of the force parallel to the displacement. Only this component does work. For example, if a force of $$10,\mathrm{N}$$ acts at $$60^\circ$$ to a displacement of $$3,\mathrm{m}$$, its parallel component is $$5,\mathrm{N}$$ and it does $$15,\mathrm{J}$$ of work.

The sign of work done

The direction of the force relative to the displacement determines the sign of the work:

• Positive work: the parallel component acts in the direction of displacement.
• Zero work: the force is perpendicular to the displacement, so it has no parallel component.
• Negative work: the parallel component acts opposite to the displacement.

Work from a force–displacement graph

A force does work when it acts through a displacement. On a force–displacement graph, the signed area between the graph and the displacement axis gives the work done over that displacement. The horizontal axis shows displacement; the vertical axis shows the component of force along the direction of displacement.

A force component in the direction of displacement is positive, so an area above the displacement axis represents positive work. A force component opposing the displacement is negative, so an area below the axis represents negative work. A force component perpendicular to the displacement does no work.

When the force varies, divide the relevant region into shapes whose areas can be calculated, or estimate the area under a curved graph. Add the areas with their signs to find the total work. If force is measured in newtons and displacement in metres, the area has units of newton metres, equivalent to joules.

Force–displacement graph: calculating signed area

Suppose a graph forms a triangle above the displacement axis from $$0$$ to $$4,\text{m}$$, with a maximum force of $$6,\text{N}$$. Its area gives positive work of $$\tfrac{1}{2} \times 4 \times 6 = 12,\text{J}$$.

From $$4$$ to $$6,\text{m}$$, suppose the graph forms a triangle below the axis, reaching $$-2,\text{N}$$. Its area contributes $$-\tfrac{1}{2} \times 2 \times 2 = -2,\text{J}$$. The total work over the full displacement is therefore $$12 - 2 = 10,\text{J}$$.

Power

Power describes how quickly work is done or energy is transferred. Transferring the same amount of energy in less time requires greater power.

The average power over a time interval is:

Here $$\Delta W$$ is the work done or energy transferred during the time interval $$\Delta t$$. The unit of power is the watt ($$\mathrm{W}$$): $$1,\mathrm{W} = 1,\mathrm{J,s^{-1}}$$. A transfer of $$1,\mathrm{J}$$ each second therefore corresponds to a power of $$1,\mathrm{W}$$.

Power from a force

When a force acts in the direction of an object's motion, it does work as the object moves. The faster the object moves, the more distance it covers each second, so the force does more work each second. For a force parallel to the velocity:

Here $$F$$ is the force and $$v$$ is the object's speed. If the force acts at an angle to the motion, use only its component in the direction of motion in place of $$F$$. Using the speed at a particular instant gives the power at that instant.

Efficiency

A device receives power as an input and transfers it to its surroundings in different ways. The useful output power is the power transferred in the way the device is intended to produce. Efficiency is the proportion of the input power transferred as useful output power.

Both powers must be measured in the same units. The units cancel in the ratio, so efficiency is dimensionless: it has no unit. To express it as a percentage, multiply the ratio by $$100$$.

For example, if a device receives $$80\ \text{W}$$ and transfers $$60\ \text{W}$$ as useful output, its efficiency is $$60/80 = 0.75$$, or $$75%$$. The remaining $$20\ \text{W}$$ is transferred in less useful ways.

A lower efficiency means that a smaller proportion of the input power becomes useful output power; a larger proportion is transferred in less useful ways.